# Let R be the region in the first and second quadrants bounded above by the graph of y=20/(1+x^2) and below by the horizontal line y=2, how do you find volume of the solid generated when R is rotated about the x-axis?

Jul 4, 2016

$= \pi \left(96 + 400 \arctan \left(3\right)\right)$

#### Explanation:

the small strip width dx as indicated will have volume when revolved about x axis given by

$\mathrm{dV} = \pi \left({y}_{2}^{2} - \pi {y}_{1}^{2}\right) \mathrm{dx}$

where ${y}_{2} = \frac{20}{1 + {x}^{2}}$ and ${y}_{1} = 2$

so

$\mathrm{dV} = \pi \left({y}_{2}^{2} - {y}_{1}^{2}\right) \mathrm{dx}$

$= \pi \left({\left(\frac{20}{1 + {x}^{2}}\right)}^{2} - {2}^{2}\right) \mathrm{dx}$

$V = \pi {\int}_{- 3}^{3} \mathrm{dx} q \quad {\left(\frac{20}{1 + {x}^{2}}\right)}^{2} - 4 \pi {\int}_{- 3}^{3} \mathrm{dx} q \quad \star$

if we do the LHS of $\star$ first as an indefinite integral just to keep it a bit cleaner:

$V = \pi \int \mathrm{dx} q \quad {\left(\frac{20}{1 + {x}^{2}}\right)}^{2}$

$x = \tan \psi , \mathrm{dx} = {\sec}^{2} \psi d \psi$

$V = 400 \pi \int \mathrm{dp} s i q \quad {\sec}^{2} \psi {\left(\frac{1}{1 + {\tan}^{2} {\psi}^{2}}\right)}^{2}$

$= 400 \pi \int \mathrm{dp} s i q \quad {\cos}^{2} \psi$

from the double angle formula $\cos 2 \psi = 2 {\cos}^{2} \psi - 1$
$= 200 \pi \int \mathrm{dp} s i q \quad \cos 2 \psi + 1$

$= 200 \pi \left(\frac{1}{2} \sin 2 \psi + \psi + C\right)$

$\sin 2 \psi = 2 \cos \psi \sin \psi$

$= 200 \pi \left(\sin \psi \cos \psi + \psi + C\right)$

see this for converting back

$= 200 \pi \left(\frac{x}{\sqrt{1 + {x}^{2}}} \cdot \frac{1}{\sqrt{1 + {x}^{2}}} + \arctan x + C\right)$

$= 200 \pi \left(\frac{x}{1 + {x}^{2}} + \arctan x + C\right)$

the RHS of $\star$ amounts to $- 4 \pi x$, so reinstating the integration interval we have

$V = \pi {\left[200 \left(\frac{x}{1 + {x}^{2}} + \arctan x\right) - 4 x\right]}_{- 3}^{3}$

$= \pi \left(96 + 400 \arctan \left(3\right)\right)$